Problem 10.4.1.
Suppose the power series \(\sum a_nx^n\) has radius of convergence \(r\) and the series \(\sum a_nr^n\) converges absolutely. Then \(\sum a_nx^n\) converges uniformly on \([-r,r]\text{.}\)
Section 10.4 Boundary Issues and Abel’s Theorem
Summarizing our results, we see that any power series \(\sum
a_nx^n\) has a radius of convergence \(r\) such that \(\sum a_nx^n\) converges absolutely when \(|x|\lt r\) and diverges when \(|x|>r\text{.}\) Furthermore, the convergence is uniform on any closed interval \([-b,b]\subset(-r,r)\) which tells us that whatever the power series converges to must be a continuous function on \((-r,r)\text{.}\) Lastly, if \(f(x)=\sum_{n=0}^\infty a_nx^n\) for \(x\in(-r,r)\text{,}\) then \(f^\prime(x)=\sum_{n=1}^\infty a_nnx^{n-1}\) for \(x\in(-r,r)\) and \(\int_{t=0}^xf(t)\dx{ t} =
\sum_{n=0}^\infty a_n\frac{x^{n+1}}{n+1}\) for \(x\in(-r,r)\text{.}\)
Thus power series are very well behaved within their interval of convergence, and our cavalier approach from Chapter 3 is justified, EXCEPT for one issue. If you go back to Problem 3.2.10 of Chapter 3, you see that we used the geometric series to obtain the series, \(\arctan x
=\sum_{n=0}^\infty(-1)^n\frac{1}{2n+1}x^{2n+1}\text{.}\) We substituted \(x=1\) into this to obtain \(\frac{\pi}{4}=\sum_{n=0}^\infty(-1)^n\frac{1}{2n+1}\text{.}\) Unfortunately, our integration was only guaranteed on a closed subinterval of the interval \((-1,1)\) where the convergence was uniform and we substituted in \(x=1\text{.}\) We “danced on the boundary” in other places as well, including when we said that
\begin{equation*}
\frac{\pi}{4}=\int_{x=0}^1\sqrt{1-x^2}\dx{x}=1+\sum_{n=1}^\infty\left(\frac{\prod_{j=0}^{n-1}\left(\frac{1}{2}-j\right)}{n!}\text{ } \right)\left(\frac{\left(-1\right)^n}{2n+1}\right)\text{.}
\end{equation*}
The fact is that for a power series \(\sum a_nx^n\) with radius of convergence \(r\text{,}\) we know what happens when \(|x|\lt r\) and when \(|x|>r\text{.}\) But we’ve never talked about what happens when \(|x|=r\text{.}\) That is because there is no systematic approach to this boundary problem. For example, consider the three series
\begin{equation*}
\sum_{n=0}^\infty x^n,\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}, \sum_{n=0}^\infty\frac{x^{n+2}}{(n+1)(n+2)}\text{.}
\end{equation*}
They are all related in that we started with the geometric series and integrated twice, thus they all have radius of convergence equal to 1. Their behavior on the boundary, i.e., when \(x=\pm 1\text{,}\) is another story. The first series diverges when \(x=\pm 1\text{,}\) the third series converges when \(x=\pm 1\text{.}\) The second series converges when \(x=-1\) and diverges when \(x=1\text{.}\)
Even with the unpredictability of a power series at the endpoints of its interval of convergence, the Weierstrass-\(M\) test does give us some hope of uniform convergence.
Unfortunately, this result doesn’t apply to the integrals we mentioned as the convergence at the endpoints is not absolute. Nonetheless, the integrations we performed in Chapter 3 are still legitimate. This is due to the following theorem by Abel which extends uniform convergence to the endpoints of the interval of convergence even if the convergence at an endpoint is only conditional. Abel did not use the term uniform convergence, as it hadn’t been defined yet, but the ideas involved are his.
Theorem 10.4.2.
Abel’s Theorem
Suppose the power series \(\sum a_nx^n\) has radius of convergence \(r\) and the series \(\sum a_nr^n\) converges. Then \(\sum a_nx^n\) converges uniformly on \([0, r]\text{.}\)
The proof of this is not intuitive, but involves a clever technique known as Abel’s Partial Summation Formula.
Lemma 10.4.3.
Let
\begin{equation*}
a_1,a_2,\,\ldots,\,a_n,\,b_1,b_2,\,\ldots\,,\,b_n
\end{equation*}
be real numbers and let \(A_m=\sum_{k=1}^ma_k\text{.}\) Then
\begin{equation*}
a_1b_1+a_2b_2+\cdots+a_nb_n=\sum_{j=1}^{n-1}A_j\left(b_j-b_{j+1}\text{ } \right)+A_nb_n\text{.}
\end{equation*}
Problem 10.4.4.
Prove Lemma 10.4.3.
Lemma 10.4.5.
Abel’s Lemma
Let \(a_1,a_2,\,\ldots,\,a_n,\,b_1,b_2,\,\ldots\,,\,b_n\) be real numbers with \(\,b_1\geq
b_2\geq\,\ldots\geq\,b_n\geq 0\) and let \(A_m=\sum_{k=1}^ma_k\text{.}\) Suppose \(|A_m|\leq B\) for all \(m\text{.}\) Then \(|\sum_{j=1}^na_jb_j|\leq B\cdot
b_1\text{.}\)
Problem 10.4.6.
Prove Lemma 10.4.5.
Problem 10.4.7.
Prove Theorem 10.4.2.
Hint.
Let \(\epsilon>0\text{.}\) Since \(\sum_{n=0}^\infty
a_nr^n\) converges then by the Cauchy Criterion, there exists \(N\) such that if \(m>n>N\) then \(\abs{\sum_{k=n+1}^ma_kr^k}\lt \frac{\epsilon}{2}\text{.}\) Let \(0\leq x\leq r\text{.}\) By Lemma 10.4.5,
\begin{equation*}
\abs{\sum_{k=n+1}^ma_kx^k}=\abs{\sum_{k=n+1}^ma_kr^k\left(\frac{x}{r}\right)^k}\leq \left(\frac{\epsilon}{2}\right)\left(\frac{x}{r}\right)^{n+1}\leq\frac{\epsilon}{2}\text{.}
\end{equation*}
Thus for \(0\leq x\leq r\text{,}\) \(n>N\text{,}\)
\begin{equation*}
\abs{\sum_{k=n+1}^\infty a_kx^k}=\lim_{m\rightarrow\infty}\abs{\sum_{k=n+1}^ma_kx^k}\leq\frac{\epsilon}{2}\lt \epsilon.\rbrack{}
\end{equation*}
Corollary 10.4.8.
Suppose the power series \(\sum a_nx^n\) has radius of convergence \(r\) and the series \(\sum
a_n\left(-r\right)^n\) converges. Then \(\sum a_nx^n\) converges uniformly on \([-r,0]\text{.}\)
Problem 10.4.9.
Prove Corollary 10.4.8.
